If `x=a(cost+tsint)` and `y=a(sint-tcost)`, find `(d^2y)/(dx^2)`

Explanation

It is given that x =a (cos t + t sin t) and
y=a (sin t - t cos t). Therefore,
`(dx)/(dt)=a[-sin t+ sin t + t cos t]= at cos t`
`(dy)/(dt)=a [ cos t -{cos t - t sin t} ] = at sin t`
`therefore" "(dy)/(dx)=(((dy)/(dt)))/(((dx)/(dt)))=(at sin t)/(at cos t)= tan t`
`"Then, "(d^(2)y)/(dx^(2))=(d)/(dx)((dy)/(dx))=(d)/(dx)(tan t)`
`=(d)/(dt) (tan t)(dt)/(dx)`
`=sec^(2) t. (dt)/(dx)`
`sec^(2)t. (1)/(at cos t)`
`(sec^(3) t)/(at)`